Practice MCQ Questions and Answer on Algebra

401.

x varies inversely as square of y. Given that y = 2 for x = 1, the value of x for y = 6 will equal to?

• (A) 3
• (B) 9
• (C) $$\frac{1}{3}$$
• (D) $$\frac{1}{9}$$

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 $$\eqalign{ & x \propto \frac{1}{{{y^2}}} \cr & \left( {{\text{Inversely proportional}}} \right) \cr & x = \frac{k}{{{y^2}}} \cr & \left( {{\text{Given}}} \right), \cr & \left( {y = 2} \right){\text{ for }}\left( {x = 1} \right) \cr & \therefore 1 = \frac{k}{{{{\left( 2 \right)}^2}}} \cr & \Rightarrow 1 = \frac{k}{4} \cr & \Rightarrow k = 4 \cr & \therefore {\text{For }}y = 6 \cr & x = \frac{4}{{{{\left( 6 \right)}^2}}} \cr & x = \frac{4}{{36}} \cr & x = \frac{1}{9} \cr} $$

402.

If $$\frac{2a+b}{a+4b}=3\text{,}$$   then find the value of $$\frac{a+b}{a+2b}=?$$

• (A) $$\frac{5}{9}$$
• (B) $$\frac{2}{7}$$
• (C) $$\frac{10}{9}$$
• (D) $$\frac{10}{7}$$

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 $$\eqalign{ & \frac{{2a + b}}{{a + 4b}} = 3{\text{ }}\left( {{\text{Given}}} \right) \cr & \Rightarrow 2a + b = 3\left( {a + 4b} \right) \cr & \Rightarrow 2a + b = 3a + 12b \cr & \Rightarrow - a = 11b \cr & \Rightarrow a = - 11b \cr & \therefore \frac{{a + b}}{{a + 2b}} \cr & \Rightarrow \frac{{ - 11b + b}}{{ - 11b + 2b}} \cr & \Rightarrow \frac{{ - 10b}}{{ - 9b}} \cr & \Rightarrow \frac{{10}}{9} \cr} $$

403.

If 1.5a = 0.04b then $$\frac{b-a}{b+a}$$   is equal to?

• (A) $$\frac{73}{77}$$
• (B) $$\frac{77}{33}$$
• (C) $$\frac{2}{75}$$
• (D) $$\frac{75}{2}$$

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 $$\eqalign{ & 1.5a = 0.04b \cr & \frac{a}{b} = \frac{{0.04}}{{1.5}} = \frac{4}{{100}} \times \frac{{10}}{{15}} = \frac{2}{{75}} \cr & {\text{Let }}a = 2x,{\text{ }}b = 75x \cr & \therefore \frac{{b - a}}{{b + a}} = \frac{{75x - 2x}}{{75x + 2x}} = \frac{{73}}{{77}} \cr & \cr & {\bf{Alternate:}} \cr & \frac{a}{b} = \frac{{0.04}}{{1.5}} \cr & \therefore \frac{{b - a}}{{b + a}} = \frac{{1.5 - 0.04}}{{1.5 + 0.04}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{1.46}}{{1.54}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{73}}{{77}} \cr} $$

404.

If x + y + z = 9, then the value of (x - 4)³ + (y - 2)³ + (z - 3)³ - 3(x - 4)(y - 2)(z - 3) is?

• (A) 6
• (B) 9
• (C) 0
• (D) 1

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 $$\eqalign{ & \mathop {\mathop {{{\left( {x - 4} \right)}^3}}\limits_ \Downarrow }\limits_{{a^3}} + \mathop {\mathop {{{\left( {y - 2} \right)}^3}}\limits_ \Downarrow }\limits_{{b^3}} + \mathop {\mathop {{{\left( {z - 3} \right)}^3}}\limits_ \Downarrow }\limits_{{c^3} - 3abc} - 3\left( {x - 4} \right)\left( {y - 2} \right)\left( {z - 3} \right) \cr & \Rightarrow a + b + c = x - 4 + y - 2 + z - 3 \cr & \Rightarrow a + b + c = x + y + z - 9 \cr & \Rightarrow a + b + c = 9 - 9 \cr & \Rightarrow a + b + c = 0 \cr} $$ $${\text{So, }}{\left( {x - 4} \right)^3} + {\left( {y - 2} \right)^3} + $$     $${\left( {z - 3} \right)^3} - $$   $$3\left( {x - 4} \right)$$  $$\left( {y - 2} \right)$$ $$\left( {z - 3} \right)$$ $$ \Rightarrow 0$$

405.

If 5x + 9y = 5 and 125x³ + 729y³ = 120, then the value of the product of x and y is?

• (A) 45
• (B) $$\frac{1}{9}$$
• (C) $$\frac{1}{135}$$
• (D) 135

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 5x + 9y = 5 . . . . . (i) 125x3 + 729y3 = 120 . . . . . (ii) From equation (i) cubing both sides ⇒ (5x + 9y)3 = 53 ⇒ 125x3 + 729y3 + 3 × 5x × 9y(5x + 9y) = 125 ⇒ 125x3 + 729y3 + 135xy × 5 = 125 ⇒ 120 + 135xy × 5 = 125 ⇒ 135xy × 5 = 5 ⇒ xy = $$\frac{1}{{135}}$$ ∴ product of x & y = $$\frac{1}{{135}}$$

406.

If $$x+\frac{1}{x+1}=1,$$    then $${\left(x+1\right)}^5$$   + $$\frac{1}{{\left(x+1\right)}^5}$$   equals?

• (A) 1
• (B) 2
• (C) 4
• (D) 8

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 $$\eqalign{ & x + \frac{1}{{x + 1}} = 1 \cr & {\text{Adding both 1 sides}} \cr & \Rightarrow x + 1 + \frac{1}{{x + 1}} = 1 + 1 \cr & \Rightarrow \left( {x + 1} \right) + \frac{1}{{\left( {x + 1} \right)}} = 2 \cr & {\text{Put }}x + 1 = 1 \cr & {\text{And }}\frac{1}{{x + 1}} = 1 \cr & \therefore {\left( {x + 1} \right)^5}{\text{ + }}\frac{1}{{{{\left( {x + 1} \right)}^5}}} \cr & = 1 + 1 \cr & = 2 \cr} $$

407.

For what value (s) of a is $$x+\frac{1}{4}\sqrt{x}+a^2$$    a perfect square?

• (A) $$\pm \frac{1}{18}$$
• (B) $$\frac{1}{8}$$
• (C) $$-\frac{1}{5}$$
• (D) $$\frac{1}{4}$$

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 $$\eqalign{ & x + \frac{1}{4}\sqrt x + {a^2} \cr & = {\left( {\sqrt x } \right)^2} + 2 \times \frac{1}{8} \times \sqrt x + {a^2} \cr & \left[ {\left( {{{\text{A}}^2} + {\text{2AB}} + {{\text{B}}^2}} \right) = {{\left( {{\text{A}} + {\text{B}}} \right)}^2}} \right] \cr & {\text{Here, A}} = \sqrt x {\text{ and }} \cr & {\text{B}} = a \cr & {\text{B}} = \frac{1}{8} \cr & \therefore a = \frac{1}{8} \cr} $$

408.

If $$3^{x+3}+7=250\text{,}$$    then x is equal to?

• (A) 5
• (B) 3
• (C) 2
• (D) 1

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 $$\eqalign{ & {3^{x + 3}} + 7 = 250 \cr & \Rightarrow {3^{x + 3}} = 250 - 7 \cr & \Rightarrow {3^{x + 3}} = 243 \cr & \Rightarrow {3^{x + 3}} = {3^5} \cr & \Rightarrow x + 3 = 5 \cr & \Rightarrow x = 2 \cr} $$

409.

If $$x-\frac{1}{x}=2\text{,}$$   then the value of the following is $$x^3-\frac{1}{x^3}$$   = ?

• (A) 2
• (B) 11
• (C) 15
• (D) 14

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 $$\eqalign{ & x - \frac{1}{x} = 2{\text{ to find }}{x^3} - \frac{1}{{{x^3}}} \cr & \Rightarrow x - \frac{1}{x} = 2 \cr & \left[ {{\text{Cubing both sides}}} \right] \cr & \Rightarrow {\left( {x - \frac{1}{x}} \right)^3} = {\left( 2 \right)^3} \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} - 3 \times x \times \frac{1}{x}\left( {x - \frac{1}{x}} \right) = 8 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} - 3 \times \left( 2 \right) = 8 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} = 14 \cr} $$

410.

If x = 2 then the value of x³ + 27x² + 243x + 631 = ?

• (A) 1321
• (B) 1233
• (C) 1231
• (D) 1211

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 $$\eqalign{ & {\text{Given, }}x = 2 \cr & {\text{Find }}{x^3} + 27{x^2} + 243x + 631 \cr & {\text{To put value }}x = 2{\text{ }} \cr & \Rightarrow {2^3} + 27 \times {2^2} + \left( {243 \times 2} \right) + 631 \cr & \Rightarrow 8 + 108 + 486 + 631 \cr & \Rightarrow 1233 \cr} $$